Step-by-Step Guide Reading Time: 15 min 4 Essential Steps

Handling Composite Functions: A Step-by-Step Guide to Domain & Range (f ∘ g)

Master the systematic approach to finding domain and range of composite functions - the most tested concept in JEE functions.

95%

JEE Papers Include

Key Steps

3-5

Marks per Question

100%

Systematic Approach

Why Composite Functions Trouble Students

Composite functions like $f(g(x))$ appear in 95% of JEE mathematics papers, yet students consistently lose marks here due to:

Confusing individual domains with composite domain
Forgetting to work from inside out
Missing the intersection step between constraints
Overlooking range compatibility between functions

🎯 The Golden Rule of Composite Functions

Domain of $f(g(x))$ = Domain of $g(x)$ ∩ $\{x: g(x) \in$ Domain of $f\}$

In simple terms: Input must work for g, AND output of g must work as input for f

🧭 Step Navigation

Step 1: Understand g(x) Step 2: Find Domain of g(x) Step 3: Apply f to g(x) Step 4: Take Intersection

Understand the Composition: f(g(x))

First, clearly identify what f(g(x)) actually means.

What is f(g(x))?

The output of g(x) becomes the input for f(x)

$$x \xrightarrow{g} g(x) \xrightarrow{f} f(g(x))$$

Visual Representation

Input → g-machine → Output₁ → f-machine → Final Output

Both machines must work for the process to succeed

📚 Example Setup

Let's use this example throughout our steps:

$f(x) = \sqrt{x}$ and $g(x) = \log(1-x)$

Then $f(g(x)) = \sqrt{\log(1-x)}$

Find Domain of g(x)

Determine all x-values that work for the inner function.

🔍 Applying to Our Example

Given: $g(x) = \log(1-x)$

Step 2.1: Logarithm condition: argument > 0

$1 - x > 0$

Step 2.2: Solve the inequality

$x < 1$

Domain of g(x): $(-\infty, 1)$

⚠️ Common Mistakes in Step 2

Forgetting that log arguments must be strictly positive
Missing denominator constraints in rational functions
Overlooking square root constraints for inner function
Not considering periodicity in trigonometric functions

Apply f to g(x) - The Range Compatibility Check

Ensure g(x) produces outputs that are valid inputs for f.

🔍 Applying to Our Example

Given: $f(x) = \sqrt{x}$ and we know $g(x) = \log(1-x)$

Step 3.1: Domain of f(x): For $\sqrt{x}$, we need $x \geq 0$

Step 3.2: Therefore, for $f(g(x))$, we need $g(x) \geq 0$

$\log(1-x) \geq 0$

Step 3.3: Solve the inequality (base 10 > 1, so inequality direction stays)

$1-x \geq 10^0 = 1$

$-x \geq 0 \Rightarrow x \leq 0$

Constraint from f: $x \leq 0$

💡 Critical Insight

This step answers: "What outputs from g are acceptable inputs for f?"

We're essentially finding: $\{x : g(x) \in \text{Domain of } f\}$

Take the Intersection - The Final Domain

Combine both constraints to find the actual domain of f(g(x)).

🔍 Applying to Our Example

From Step 2: Domain of g(x): $x < 1$

From Step 3: Constraint from f: $x \leq 0$

Step 4.1: Take intersection of both conditions

$\{x < 1\} \cap \{x \leq 0\} = \{x \leq 0\}$

🎉 Final Answer

Domain of $f(g(x)) = (-\infty, 0]$

⚠️ The Intersection Trap

Students often make these mistakes in Step 4:

Taking union instead of intersection
Forgetting that both conditions must be satisfied simultaneously
Missing edge cases where one constraint is stricter than the other

Finding Range of f(g(x))

Step-by-Step Process

Find range of g(x) over its domain
Take intersection with domain of f(x)
Apply f to this restricted set
The result is the range of f(g(x))

Mathematical Form

$$ \text{Range of } f(g(x)) = f(\text{Range of } g(x) \cap \text{Domain of } f(x)) $$

📚 Range Example

Given: $f(x) = \sqrt{x}$, $g(x) = \log(1-x)$, Domain: $x \leq 0$

Step 1: Range of g(x) for $x \leq 0$:

When $x \leq 0$, $1-x \geq 1$, so $\log(1-x) \geq 0$

As $x \to -\infty$, $g(x) \to +\infty$

Range of g: $[0, \infty)$

Step 2: Intersection with domain of f ($x \geq 0$): $[0, \infty)$

Step 3: Apply f: $f([0, \infty)) = [0, \infty)$

Range of f(g(x)): $[0, \infty)$

Advanced Composite Function Examples

Example 1: Triple Composition

Find domain of $h(g(f(x)))$ where:

$f(x) = \sqrt{x}, \quad g(x) = \frac{1}{x-1}, \quad h(x) = \log(x)$

Solution Approach

Step 1: $f(x) = \sqrt{x}$ → Domain: $x \geq 0$

Step 2: $g(f(x)) = \frac{1}{\sqrt{x}-1}$ → Need $\sqrt{x} \neq 1$ → $x \neq 1$

Step 3: $h(g(f(x))) = \log\left(\frac{1}{\sqrt{x}-1}\right)$ → Need $\frac{1}{\sqrt{x}-1} > 0$

Step 4: Solve $\frac{1}{\sqrt{x}-1} > 0$ → $\sqrt{x}-1 > 0$ → $x > 1$

Step 5: Intersection: $x \geq 0$ and $x \neq 1$ and $x > 1$ → $x > 1$

Domain: $(1, \infty)$

Example 2: Trigonometric Composite

Find domain of $f(g(x))$ where:

$f(x) = \sqrt{\sin^{-1}x}, \quad g(x) = \frac{x-1}{x+1}$

Solution Approach

Step 1: Domain of g(x): $x \neq -1$

Step 2: Domain of f(x): For $\sqrt{\sin^{-1}x}$, need $\sin^{-1}x \geq 0$ and $-1 \leq x \leq 1$

Step 3: $\sin^{-1}x \geq 0$ when $0 \leq x \leq 1$

Step 4: So for f(g(x)), need $0 \leq g(x) \leq 1$

Step 5: Solve $0 \leq \frac{x-1}{x+1} \leq 1$ with $x \neq -1$

📋 Quick Reference Guide

Domain Rules

Work inside out - start with innermost function
Take intersections - not unions
Check range compatibility between functions
Remember edge cases and boundary values

Common Patterns

$\sqrt{g(x)}$ → $g(x) \geq 0$
$\log(g(x))$ → $g(x) > 0$
$\frac{1}{g(x)}$ → $g(x) \neq 0$
$\sin^{-1}(g(x))$ → $-1 \leq g(x) \leq 1$

🎯 Practice Problems

Apply the 4-step method to these problems:

1. Find domain of $f(g(x))$ where $f(x) = \sqrt{x-2}$, $g(x) = \log(x+1)$

Hint: Apply both logarithm and square root constraints

2. Find domain and range of $f(g(x))$ where $f(x) = \frac{1}{x}$, $g(x) = x^2-4$

Hint: Watch for denominator becoming zero

3. Find domain of $g(f(x))$ where $f(x) = \sqrt{9-x^2}$, $g(x) = \log(x-1)$

Hint: Note the order changed to g(f(x))

Mastered Composite Functions?

You now have a systematic approach to tackle any composite function problem in JEE